Problem: Simplify and expand the following expression: $ \dfrac{4}{n + 9}+ \dfrac{3}{3n + 18}+ \dfrac{1}{n^2 + 15n + 54} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{3}{3n + 18} = \dfrac{3}{3(n + 6)}$ We can factor the quadratic in the third term: $ \dfrac{1}{n^2 + 15n + 54} = \dfrac{1}{(n + 9)(n + 6)}$ Now we have: $ \dfrac{4}{n + 9}+ \dfrac{3}{3(n + 6)}+ \dfrac{1}{(n + 9)(n + 6)} $ The least common multiple of the denominators is: $ (n + 9)(n + 6)$ In order to get the first term over $(n + 9)(n + 6)$ , multiply by $\dfrac{3(n + 6)}{3(n + 6)}$ $ \dfrac{4}{n + 9} \times \dfrac{3(n + 6)}{3(n + 6)} = \dfrac{12(n + 6)}{(n + 9)(n + 6)} $ In order to get the second term over $(n + 9)(n + 6)$ , multiply by $\dfrac{n + 9}{n + 9}$ $ \dfrac{3}{3(n + 6)} \times \dfrac{n + 9}{n + 9} = \dfrac{3(n + 9)}{(n + 9)(n + 6)} $ In order to get the third term over $(n + 9)(n + 6)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{1}{(n + 9)(n + 6)} \times \dfrac{3}{3} = \dfrac{3}{(n + 9)(n + 6)} $ Now we have: $ \dfrac{12(n + 6)}{(n + 9)(n + 6)} + \dfrac{3(n + 9)}{(n + 9)(n + 6)} + \dfrac{3}{(n + 9)(n + 6)} $ $ = \dfrac{ 12(n + 6) + 3(n + 9) + 3} {(n + 9)(n + 6)} $ Expand: $ = \dfrac{12n + 72 + 3n + 27 + 3}{3n^2 + 45n + 162} $ $ = \dfrac{15n + 102}{3n^2 + 45n + 162}$ Simplify: $ = \dfrac{5n + 34}{n^2 + 15n + 54}$